Applied
Question 6:
Consider the simple function R(β) = sin(β) + β/10
- Question 6-a: Draw a graph of this funciton over
the range β ∈ [-6, 6]
- Answer:
- Answer:
- Question 6-b: What is the derivative of this
function?
- Answer:
The derivative of the funciton R(β) = sin(β) + β/10 is δR(β) = cos(β).- Question 6-c:
- Answer:
[1] -0.5337573
- Question 6-d:
- Answer:
[1] -1.15708
Question 7:
Fit a neural network to the Default data. Use a single hidden layer with 10 units, and dropout regularization. Have a look at Labs 10.9.1-10.9.2 for guidance. Compare the classification performance of your model with that of linear logistic regression.
Model: "sequential"
________________________________________________________________________________
Layer (type) Output Shape Param #
================================================================================
dense_1 (Dense) (None, 10) 50
dropout (Dropout) (None, 10) 0
dense (Dense) (None, 1) 11
================================================================================
Total params: 61 (244.00 Byte)
Trainable params: 61 (244.00 Byte)
Non-trainable params: 0 (0.00 Byte)
________________________________________________________________________________105/105 - 0s - 182ms/epoch - 2ms/stepThe accuracy of the neural network model is 96.640%. truth
prediction 0 1
0 3215 108
1 4 6The accuracy of the logistic regression model is 97.420%. prediction
truth 0 1
0 3209 10
1 76 38Question 8:
From your collection of personal photographs, pick 10 images of animals (such as dogs, cats, birds, farm animals, etc.). If the subject does not occupy a reasonable part of the image, then crop the image. Now use a pretrained image classification CNN as in Lab 10.9.4 to predict the class of each of your images, and report the probabilities for the top five predicted classes for each image.
Model: "sequential"
________________________________________________________________________________
Layer (type) Output Shape Param #
================================================================================
dense_1 (Dense) (None, 10) 50
dropout (Dropout) (None, 10) 0
dense (Dense) (None, 1) 11
================================================================================
Total params: 61 (244.00 Byte)
Trainable params: 61 (244.00 Byte)
Non-trainable params: 0 (0.00 Byte)
________________________________________________________________________________1/1 - 1s - 829ms/epoch - 829ms/step
The predicted classes and probabilities of the images of dogs and cats are as
follows with respect to order of appearance:$birds.jpeg
class_name class_description score
1 n01820546 lorikeet 0.961650610
2 n01818515 macaw 0.009002453
3 n01806143 peacock 0.008250927
4 n03874599 padlock 0.005620250
5 n02999410 chain 0.002654100
$butterfly.jpeg
class_name class_description score
1 n04423845 thimble 0.58907545
2 n04409515 tennis_ball 0.17006825
3 n03958227 plastic_bag 0.03801656
4 n03871628 packet 0.01514976
5 n01773549 barn_spider 0.01233464
$cat1.jpeg
class_name class_description score
1 n02124075 Egyptian_cat 0.20266204
2 n04265275 space_heater 0.13892646
3 n02105412 kelpie 0.10062172
4 n02104365 schipperke 0.09486673
5 n02110806 basenji 0.06643521
$cat2.jpeg
class_name class_description score
1 n03223299 doormat 0.21453752
2 n04074963 remote_control 0.09869209
3 n02113186 Cardigan 0.07454871
4 n02123045 tabby 0.06722851
5 n02123159 tiger_cat 0.06283280
$cat3.jpeg
class_name class_description score
1 n02971356 carton 0.9914653897
2 n04004767 printer 0.0020149811
3 n03127925 crate 0.0016998923
4 n02916936 bulletproof_vest 0.0009906454
5 n02105162 malinois 0.0003613399
$dog1.jpg
class_name class_description score
1 n02098413 Lhasa 0.484432071
2 n02086240 Shih-Tzu 0.462051690
3 n02097474 Tibetan_terrier 0.016058447
4 n02113624 toy_poodle 0.008868926
5 n02094433 Yorkshire_terrier 0.007705405
$dog2.jpeg
class_name class_description score
1 n02108551 Tibetan_mastiff 0.29876935
2 n02111277 Newfoundland 0.29656842
3 n02106382 Bouvier_des_Flandres 0.26543206
4 n02112137 chow 0.07458649
5 n02105056 groenendael 0.03799465
$dog3.jpeg
class_name class_description score
1 n02086240 Shih-Tzu 0.88302189
2 n02097047 miniature_schnauzer 0.04248711
3 n02110627 affenpinscher 0.02213948
4 n02085936 Maltese_dog 0.01723721
5 n02098413 Lhasa 0.01470254
$dog4.jpeg
class_name class_description score
1 n02086240 Shih-Tzu 0.60574049
2 n02098413 Lhasa 0.15268034
3 n02110627 affenpinscher 0.07483345
4 n02085936 Maltese_dog 0.05702168
5 n02094433 Yorkshire_terrier 0.01930033
$dog5.jpeg
class_name class_description score
1 n02086240 Shih-Tzu 0.824865639
2 n02098413 Lhasa 0.127952948
3 n02110627 affenpinscher 0.024699720
4 n02113624 toy_poodle 0.011461542
5 n02086079 Pekinese 0.003910229Question 9:
Fit a lag-5 autoregressive model to the NYSE data, as described in the text and Lab 10.9.6. Refit the model with a 12-level factor representing the month. Does this factor improve the performance of the model?
[1] "dayfri" "daymon" "daythur"
[4] "daytues" "daywed" "L1.DJ_return"
[7] "L1.log_volume" "L1.log_volatility" "L2.DJ_return"
[10] "L2.log_volume" "L2.log_volatility" "L3.DJ_return"
[13] "L3.log_volume" "L3.log_volatility" "L4.DJ_return"
[16] "L4.log_volume" "L4.log_volatility" "L5.DJ_return"
[19] "L5.log_volume" "L5.log_volatility"
56/56 - 0s - 241ms/epoch - 4ms/stepThe R-squared value of the lag-5 autoregressive model with a 5-level factor
representing the day of the week is 0.453.
56/56 - 0s - 103ms/epoch - 2ms/stepThe lag-5 autoregressive model refit with a 12-level factor representing the
month has an R-squared value of 0.413.The inclusion of the 12-level factor representing the month reduced the fit of
the model to the data.Question 10:
In Section 10.9.6, we showed how to fit a linear AR model to the NYSE data using the lm() function. However, we also mentioned that we can “flatten” the short sequences produced for the RNN model in order to fit a linear AR model. Use this latter approach to fit a linear AR model to the NYSE data. Compare the test R-squared of this linear AR model to that of the linear AR model that we fit in the lab. What are the advantages/disadvantages of each approach?
[1] 6046 5 356/56 - 0s - 114ms/epoch - 2ms/stepThe R-squared value of the linear autoregressive model fit in the lab is:
0.413223. The R-squared value of the linear autoregressive model fit using
a flattened sequences produced for the RNN model is 0.411252. These figures
are identical to at least 5 significant digits. The advantages of using the
lm function to create an autoregressive model is a reduction in time training
the model. The advantage of using a neural network is the ability to add
non-linearities to the existing model by including hidden layers in the future.Question 11:
Repeat the previous exercise, but now fit a nonlinear AR model by “flattening” the short sequences produced for the RNN model.
56/56 - 0s - 132ms/epoch - 2ms/stepThe R-squared value of the flat-input, non-linear autoregressive model is
0.41140.Question 12:
Consider the RNN fit to the NYSE data in Section 10.9.6. Modify the code to allow inclusion of the variable day_of_week, and fit the RNN. Compute the test R-squared.
56/56 - 1s - 522ms/epoch - 9ms/stepThe test R-squared of the RNN fit to include the day of the week is 0.4709660.Question 13:
Repeat the analysis of Lab 10.9.5 on the IMDb data using a similarly structured neural network. There we used a dictionary of size 10,000. Consider the effects of varying the dictionary size. Try the values 1000, 3000, 5000, and 10,000, and compare the results.
The model training accuracy for dictionary size of 1000:
Final epoch (plot to see history):
loss: 0.3031
accuracy: 0.8773
val_loss: 0.3461
val_accuracy: 0.86
The model test accuracy for dictionary size of 1000:
Final epoch (plot to see history):
loss: 0.3032
accuracy: 0.8779
val_loss: 0.3321
val_accuracy: 0.8598
The model training accuracy for dictionary size of 3000:
Final epoch (plot to see history):
loss: 0.2003
accuracy: 0.9278
val_loss: 0.3678
val_accuracy: 0.8667
The model test accuracy for dictionary size of 3000:
Final epoch (plot to see history):
loss: 0.2003
accuracy: 0.9286
val_loss: 0.3559
val_accuracy: 0.8676
The model training accuracy for dictionary size of 5000:
Final epoch (plot to see history):
loss: 0.138
accuracy: 0.9551
val_loss: 0.4105
val_accuracy: 0.86
The model test accuracy for dictionary size of 5000:
Final epoch (plot to see history):
loss: 0.139
accuracy: 0.9552
val_loss: 0.4844
val_accuracy: 0.8555
The model training accuracy for dictionary size of 10000:
Final epoch (plot to see history):
loss: 0.03823
accuracy: 0.9917
val_loss: 0.707
val_accuracy: 0.864
The model test accuracy for dictionary size of 10000:
Final epoch (plot to see history):
loss: 0.04653
accuracy: 0.9922
val_loss: 0.7714
val_accuracy: 0.8451 The training accuracy of the model increased with dictionary size. The test
validation accuracy peaked at a dictionary size of 3000. Increasing the
dicitonary size did not necessarily increase the test validation accuracy.